cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A156924 Fifth right hand column (n-m=4) of the A156920 triangle.

Original entry on oeis.org

1, 83, 2685, 56285, 919615, 12813843, 160206627, 1854550395, 20291056470, 212826091180, 2161547322134, 21414479565774, 208076662576370, 1991164206775450, 18825064380813450
Offset: 0

Views

Author

Johannes W. Meijer, Feb 20 2009

Keywords

Crossrefs

Other columns A000340, A156922, A156923.
Equals A156920 fifth right hand column.
Equals A156919 fifth right hand column divided by 16.
Equals A142963 fifth right hand column divided by 2^n

Formula

a(n)=55*a(n-1)-1365*a(n-2)+20251*a(n-3)-200557*a(n-4)+1402203*a(n-5)-7137473*a(n-6)+26886431*a(n-7)-75433971*a(n-8)+157376597*a(n-9)-241846607*a(n-10)+268663713*a(n-11)-208880991*a(n-12)+107416665*a(n-13)-32730075*a(n-14)+4465125*a(n-15)
a(n)= (16*n^4-7776*n^3*3^n+256*n^3-104976*n^2*3^n+225000*n^2*5^n+ 1496*n^2- 464616*n*3^n+ 2250000*n*5^n-2016840*n*7^n+3776*n-673596*3^n+5568750*5^n-11092620*7^n+6200145*9^n+3465)/6144
G.f.: GF1(z;RHCnr=5) = (1+28*z-515*z^2+1654*z^3+8689*z^4-65864*z^5+142371*z^6-82242*z^7-99090*z^8+113400*z^9)/((1-9*z)*(1-7*z)^2*(1-5*z)^3*(1-3*z)^4*(1-z)^5)

A142964 a(n) = 6*2^n - 2*n - 5.

Original entry on oeis.org

1, 5, 15, 37, 83, 177, 367, 749, 1515, 3049, 6119, 12261, 24547, 49121, 98271, 196573, 393179, 786393, 1572823, 3145685, 6291411, 12582865, 25165775, 50331597, 100663243, 201326537, 402653127, 805306309, 1610612675, 3221225409, 6442450879, 12884901821
Offset: 0

Views

Author

Wolfdieter Lang, Sep 15 2008

Keywords

Comments

Previous name was: One half of second column (m=1) of triangle A142963.
Essentially a duplicate of A050488. - Johannes W. Meijer, Feb 20 2009

Examples

			a(3) = 6*2^3 - 2*3 - 5 = 37.

References

  • Eric Billault, Walter Damin, Robert Ferréol, and Rodolphe Garin, MPSI Classes Prépas - Khôlles de Maths, Exercices corrigés, Ellipses, 2012, exercice 2.22 (1) pp 26, 43-44.

Links

Crossrefs

Cf. A142965 (m=2 column/4).
Equals A050488(n+1).
Equals A156920(n+1,1).
Equals A156919(n+1,1)/2^n.
Cf. A156925, A339771.
Partial sums of A033484.

Programs

  • Maple
    seq(6*2^n-2*n-5,n=0..40); # Bernard Schott, Dec 16 2020
  • Mathematica
    a[n_]:=6*2^n-2n-5;Array[a,32,0] (* or *) CoefficientList[Series[(1+x)/((1-x)^2*(1-2*x)),{x,0,31}],x] (* or *) LinearRecurrence[{4,-5,2},{1,5,15},32] (* James C. McMahon, Aug 12 2025 *)
  • PARI
    Vec((1+z)/((1-z)^2*(1-2*z)) + O(z^50)) \\ Michel Marcus, Jun 18 2017

Formula

a(n) = A142693(n+2,1)/2.
From Johannes W. Meijer, Feb 20 2009: (Start)
a(n) = 4a(n-1) - 5a(n-2) + 2a(n-3) for n > 2 with a(0) = 1, a(1) = 5, a(2) = 15.
G.f.: (1+z)/((1-z)^2*(1-2*z)). (End)
a(n) = Sum_{i=0..n} Sum_{j=0..n} 2^min(i,j) (Billault et al) (compare with A339771 that has max instead of min). - Bernard Schott, Dec 16 2020
a(n) = 2*A066524(n+1) - A339771(n). - Kevin Ryde, Dec 17 2020
E.g.f.: 6*exp(2*x) - exp(x)*(5 + 2*x). - Stefano Spezia, Dec 17 2020

Extensions

New name using a formula of Bernard Schott by Peter Luschny, Dec 17 2020

A142962 Scaled convolution of (n^3)*A000984(n) with A000984(n).

Original entry on oeis.org

4, 26, 81, 184, 350, 594, 931, 1376, 1944, 2650, 3509, 4536, 5746, 7154, 8775, 10624, 12716, 15066, 17689, 20600, 23814, 27346, 31211, 35424, 40000, 44954, 50301, 56056, 62234, 68850, 75919, 83456, 91476, 99994, 109025, 118584, 128686, 139346, 150579
Offset: 1

Views

Author

Wolfdieter Lang, Sep 15 2008

Keywords

Comments

S(3,n) := Sum_{j=0..n} j^3*binomial(2*j,j)*binomial(2*(n-j),n-j).
a(n) = 2^3*S(3,n)/4^n, n >= 1.
O.g.f. for S(3,n) is G(k=3,x). See triangle A142963 for the general G(k,x) formula.
The author was led to compute such sums by a question asked by M. Greiter, Jun 27 2008.

Links

Crossrefs

Cf. A142961 triangle: row k=3: [3, 5], with the row polynomial 3+5*n.
Cf. A049451 (scaled k=2 case).

Programs

  • Mathematica
    Rest@ CoefficientList[Series[x (4 + 10 x + x^2)/(1 - x)^4, {x, 0, 39}], x] (* Michael De Vlieger, Jul 02 2023 *)

Formula

a(n) = n^2*(3+5*n)/2.
a(n) = (2^3)*S(3,n)/4^n with the convolution S(3,n) defined above.
G.f.: x*(4+10*x+x^2)/(1-x)^4. - Joerg Arndt, Jul 02 2023

A211608 Triangle T(n,k), 0 <= k <= n, given by (0, 1, 0, 2, 0, 3, 0, 4, 0, 5, ...) DELTA (1, 2, 3, 4, 5, 6, 7, 8, 9, ...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 1, 9, 15, 0, 1, 21, 90, 105, 0, 1, 45, 375, 1050, 945, 0, 1, 93, 1350, 6825, 14175, 10395, 0, 1, 189, 4515, 36750, 132300, 218295, 135135, 0, 1, 381, 14490, 178605, 992250, 2765070, 3783780, 2027025
Offset: 0

Views

Author

Philippe Deléham, Feb 10 2013

Keywords

Examples

			Triangle begins :
1
0, 1
0, 1,  3
0, 1,  9,   15
0, 1, 21,   90,  105
0, 1, 45,  375, 1050,   945
0, 1, 93, 1350, 6825, 14175, 10395

Links

Crossrefs

Cf. A000079, A001147, A048993, A187075, A211402, A142963

Formula

T(n,k) = A048993(n,k)*A001147(k).
T(n,k) = A211402(n,k)/(2^(n-k)).
T(n,k) = k*T(n-1,k) + (2*k-1)*T(n-1,k-1), T(0,0) = 1, T(n,k) = 0 if k<0 or if k>n.
G.f.: F(x,t) = 1 + x*t + (x+3*x^2)*t^2/2! + (x+9*x^2+15*x^3)*t^3/3! + ... = Sum_{n = 0..inf} R(n,x)* t^n/n!.
The row polynomials R(n,x) satisfy the recursion R(n+1,x) = (x+2*x^2)*R'(n,x) + x*R(n,x) where ' indicates differentiation with respect to x.
R(n,x) = 1/sqrt(1 + 2*x)*Sum_{k >= 0} binomial(2*k,k)/2^k*k^n * x^k/(1 + 2*x)^k (see Boyadzhiev, eqn. 19). - Peter Bala, Jan 18 2018
Previous Showing 11-14 of 14 results.

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